# Building an algebraic closure

I need to show you how you can add the root of an element in an overly complicated and abstract way. If you’re thinking of $\mathbb Q$, or something like that, and you want to add a root of $x^2-2$, you can just tack on $\sqrt2$ to make $\mathbb Q(\sqrt2)$. But what if you don’t know a priori what the root is that you want to add? Or what if the field isn’t something you understand well? What if it’s one of those confusing finite fields?

Disclaimer: I’m going to use a fair amount of ring theory without even really defining the terms. If you feel lost, don’t worry about it. I’ve included links so you can try to learn what you want. This sort of thing will be restricted to this post, so if you don’t understand, you’ll still be fine for later posts.

Let’s try to be very general. $k$ is a field, and $p$ is an irreducible polynomial (meaning it can’t be factored) with coefficients in $k$. That is, $p\in k[x]$. What I can do is let $\alpha$ be a variable, and consider the ring $k[\alpha]$. This is exactly the same as $k[x]$ (just replacing $x$ with $\alpha$). The reason I switch is because I don’t want to confuse the two. I want $x$ to always represent a variable. In a moment, $\alpha$ will be the root of our polynomial.

Now I’m going to take the ideal generated by $p(\alpha)$ in $k[\alpha]$. We write this as $(p(\alpha))$ for reasons I don’t totally understand. In any event, this is a prime ideal. That means that if $Q\cdot R$ is in the ideal, then one of $Q$ and $R$ are as well. The reason I know this is because we chose $p$ to be irreducible. The analogous situation in the integers is that if $q\cdot r$ is some multiple of a prime $p$, then at least one of $q$ and $r$ are divisible by this prime.

When I mod out by this ideal $(p(\alpha))$, because it was a prime ideal, and therefore maximal (since $k[x]$ is a principal ideal domain), I get a field. That’s a consequence of the correspondence theorem. Sweet.

But what’s even sweeter is that in this ring, $\alpha$ is a root of the polynomial $p(x)$. Since modding out by $(p(\alpha))$ essentially says “I declare that $p(\alpha)$ is zero!”

What’s even even sweeter is that $k$ is a subfield of this new field $k[\alpha]/(p(\alpha))$.

This whole thing was a bit nasty, but it means that we can take any field, and add the root of a polynomial, even if we don’t know a priori what that root is.